i,:ep Functions
.3
iLEMS
.In o-^L
.,:;,,
^r n
.,,i, I through (r, skctch
l.; g(l) ::':."
= trt(t) l2Lr.t(t) _ r.tttr(t)
g(r) = f (t
-5.
- l)tt2(t).
u,here.l.(t)
rhe gr.aph of the
-qiven funcrion on the ilnIeIVal
2. gt,t) = (t
-
erc f
=
I
(.t)
2/
-
Jlll
_ L)ut(t) _2(t _ 2)th(t)t
(/ _ -3)u,(r)
6. g(l ) = (t
In each of problenrs 7 through
i2:
(a) Sketch the gr.aph of thcr
given function.
(b) Explessl (/) in ter.nts
of the unir step function
0. [) <r
f
t-
7. f
rt7
f r. 0<r<1,
<3.
l-2, 3<r <
(I z. .5<r <7.
8 rft):l ;
t-j'
.5,
=
{ t. t>7.
q. t'1t
=
11'
(t 2r'
lc
I
o 5 t<1.
l>
2.
< l,
11. .f
r).
r/,
-1
.?
(t) =
,=1.;,
'o
lo. 1i1v=[t,2' o<t<2'
"-=j
11, t>2.
oar<2'
t2 ,,,,:
fl'
ll _, :: l. ;
1,,. ,.,
In each of problems J 3
through lB. find the Laplace
transfor.m of the given lunction.
13. rrr)
=
r <2
/o'
(a\
I-5,
t<1
r+.i6t=/0.
+2.
>l
I
It'-2t
[(r-2)r. r>2
ltl=Jr-|
t^
[''.
/.:r
"'
' :l<2t
16.
t 7 2n
17.
f(t) : (t -3)Ltzft.) -
1t).
F(.s)-
(t
fO=
rrr
(r) -r 2utlrJ
-
_2)14(r)
bule)
1g. .f(/) = t _ ttt(r)(r_ l),
I>0
ln each of problems 19 through
24,find, the inverse Laplace
transform of the given function.
3l
(s
_
20. Fk)
2)o
1,.,
rl
', -i,1lC
-tr/tr,.'),=__;_-_
-,.
.i"_/.!*2
23. Fr,r) _ l.t-2)e'
,r'. - 4.r t- 3
25. Suppose that F(s)
(a)
- -2.t
__'=_- .i,rJ-,2
It
._.
r-(.lt = __
.i.-4
--;----
2"i. f'1r1
= L{.f (r.1} exists for.s
Shorv that if c is u positirc.o,lrto,.,t.
>n>
:
e
+e.!\-C-1r-(,.1\
(J.
th"n'
L{r(ct)t:Irf:) ,
,>cLt
> 0.
3)ttz(t) _ (t _ 2.)Lt1(t)
/
334
Chapter 6. The Laplace Trat
(b) Shorv tl.rat if /r is a positive constant, then
t'
=
lFrk.:,,
(c) Shou, that if c and b are constants with n
L-11Flas
+ b)l
=
(
I,
>
O)
0, then
t;rr",,
(i,)
In each of Problems 26 through 29, use the results of Problem
trirnsforrn oI the given lunction
1n*) yl
F(, = 4{/rrl
4u r..\'r/ 9e 26,
1
+J
In each of Problems 30 through
)'7
Fr, r
?O
Fi.r-'-
2-5
to find the ipvcr..
?t I
-,
1
'-1f+.4i+5
,'2 o-4s
2s-1
12s
33, find the Laplace transfor.m of the given fu:.-._
Ploblem 33, assunte that term-by-term integration of the infinite series is permissib,.:
[l
o:l
:u
'" l,rr:
"'-lo.l1' r>1 '
t
r,,,: .|u'
,1.
'''"':1,.
o<t<l
t <r <2
2sr<3
Io' r>3
rrrrr
32 t''@ =1-ur(r)+..,!Lt2,,G)-u2n+t (r)= L+f (-1)klr(/)
33
f
(tl: 1 * I (-i)trrr(/).
See Frgule 6.3,7
FIGURE 6.3.7 The function.l (r) in Problem
33; a square wave.
34 Letl satisfy t( +T):11r) for all r > 0 and for some fixed positive number z
to be periodic rvith period 7 on 0 < r < co. Show that
[1'
at f ,,,
L\.i tt tl
In each of Problems 35 through
of the given function.
-1,), /(/)-
I
{
l. 'r-
1
< l.
[,,. I-r-2:
.f tt +21 - 1,,,
Compare rvith Ploblem 33.
38. use the
l
Jt)
e-',
f
dt
(t1
| - e-'T
result of Problem 34 to find the Lapiace tr:
.16,
rrr,:
I
l-1.
(- | . t .
I
|(t-rz1:t14
See Figure 6.3.8.
335
,Srsr Funcfions
FIGURE 6.3.8 'I'he iunction f (r) in Problem
31.1(():r.
0<1<11
3ii..ltt)
36: a square wave
=sinl,
0
.f(t+L):fO.
f (r
Seo Figure 6.3,9.
See Figure 6.3.10.
+
r'1
=
</ <;r;
1111.
v
I
FIGURE 6.3.9
-fhe
function./
t)
FIC;URE 6.3.10 ]he function.l u) in
in
Plohlern 38; a rectifiecl sine rvavc.
Problem 37: a sarvtooth rvave.
If /(r):
\,:.f (t).
39. (a)
(b) Let B(/) :
1
I'
I
J,t
uq(r), find
comparc with Problent 30. Skctch the graph ot
I{frr)}
/te trtt. where the functio'
1,
:
gtr)
),
:
h(t) and frnd L{h(t)\.
/
is definccl
in part (a), Sketch the graph of
l{g(t)}.
(c) Letl(1):g(rl *u)tt)g(t --1).wheregisdelineclinpart(b) Sketchthegraphof
ancl hnd
40. Consicler the functionp defined by
(t. o < t
P,,,:1r_, f ., I I,
t(a) Sketch the graph sf -u : p(r).
(b) Fincl L\ptt)l b1'noting rhzrt
7.r
+2):
p(t)
is the perodic cxtcusion of the flrnctron lt
Problcnr .lt)(c) and then using thc result of Prol..lenl
(c) Find L\p(rt1
pu
3'1.
by noting that
t'
1.r1t- Ja| lrltrlt.
rvltcre ./ is the f unction in Ploblenr 36, and then using Theoretlt 6.2.I
.
in
340
Chapter 6. The Laplace Trz
li
' i,
r
FIGURE 6.4.3 Solution o1 the initial value ploblem (16), (17)
r:
Note that in this exanrple. the forci's function g is continuous
but g, is discor.
-5 and r = r0. It follows that the solutior.r q) ancl its
flrsr trvo
evervwhere. but @"' has discontinuities at r
g' at those points.
:
5 and
at, :
derivatives are ,
10 that match the discor::
(
PROBLEMS
In each of Problems I through L3:
(a) Find the solution of the given inrtial value pt.oblent,
(b) Drari'the graphs ol the solution ancl of the forcr'g funcrion;
explain ho*, ther
--rry!*=:es3
6'Z 7,y"+-v=/(r); -y(0):0, l,(0):r; t,r :l],' 01t<3n
[u. rn:r<\
6e
{a
+2y:7111'
2.
)"' +21,'
3.
1"'*,1_r,: sirrl
+ 4.\'= sini
6a'
6a5
4. .),"
{a6
t"*3r,'
+21'
*
/(0)
:0,
*2r):
rr"(r)sin(r - r)l
trz,(/)sin(.t
= l'(r1:
r'(0)
:0.
-y'(0)
=
1.
:61. 1,(0t :0
.),(0) :0. ),,(t)) : u
y,(i))
:0;
6aE
dt)
,"1.1':,g(1):
6aj
e
*
(r <i < 1(l
fi,l : ft
l(1, r'l()
1
.i(0)
=(1.
,u',(01
=1;
.r,,(0)
(^
g(r):l:,,
t-
d?
ro )r'T,\,'-r-
J.r,=g(r)t
:0,
=0;
6), tt v" + 4l : un(t) * hft(.t); y(0) : 0, ]'(0) :
r:,;i.fi,
: It, n < | < 2t
[U. U<r<:r ani
l(0)
3r" +2t - the). .1,(0) = 0.
),,(0) :
:
*
rr:.(t);
:
l,
.1"'
1, l,'(0) : 0
r,(0)
r'''+,!"+ lt,=l -rr":(i)(/ - :rl2): .v(0) =0,
,),"
1,t,,
l,(0)
1'(0)
*,r,
0
:11
:'r=j'o
: ft"t'.
[u.
t) < | < t
| > tr
l: Lf qyg49 " t -!t hp
con tin u o u
s Fo
rc
in
g Fu n c t io n s
341
* ), : Ltt() - u2(j); y(0) : 0,
.t'(0):0. .y"(0) :0, )""(0) = 0
5J'"*4y
*
:
l u.(t); l(0) : 0. ,r,'(0) = 0. ,r,'(O) : 0^ 1,"'(Q) = 0
)i4r
f
12. ),1t)
|'
l-1.
t'
is
14.
Frnd an expression inr.'olvir-rg u. (t) for'a function.f that rantps up from
zero at I
value /r at / : to * k.
l -5.
f lnct an expression involving ti.(t) for a function g that ranrps up 1r.om zero
at
valuc /r at t : tu* k and then ranrps back clou,n to zero at t : tt 2k.
*
16.
1
:
=
to
to the
1o
to the
A certarn spring-ntass svsterr satrsfies the initial value problent
11
n'here s(r)
:
r13,2(.t)'
(0)
=
0'
r.,,lll;.'
,),.',',,,,;':;,if
(a) Sketch the graph of g(ll. Observe that
':, it
One ltrne unit.
is a pulse of unit magnitude extelding oyer.
(b) Solve the initiai value problerr.
(c) Plot the solution for /< : 112,lt = l. and k :
solution and horv they depend on k.
2. Describe the
principal featur.es of the
(d) Find. to trvo decimal places. the smallest value of k for rvhich the solution u(l) reaches
the value 2,
i'
(e) Suppose /r : 2, Fincl lhc tirre r afreru,hich Iu(l)l < 0,1 for all r > r,
17. Modify'the ploblem in Example 2 of this sectton by, replacing the given
forcing function
g(I.) by
l(/) :
[15(r)(r -.5)
-
rr5*1(1)(.t
,5 - b] lk.
(a) Sketch the graph of f (tl anc.l descrrbe horv it depends on k. For rvhat yalue of k is (l)
f
identical to g(l) in the exantple?
(b) Solve the initial valuc problent
,v,
1{1,
-.1'(r),
_t,(0)
:0.
}'(0)
,-i
:0.
(c) The solutiorl in part (b) depends on k. but tor sufficientlv large r the solution is airvavs
a simple harntonic oscillation about,r,= 1/,{. Tr},,to decicje hctrv the amplitude of this
evelltual osciliation depends on /r. Then confirr.n your conclusion by plotting the solution
tor
t'
a ferv
different values oi'/t.
i
18. Consider the initiai value problerr
:.
y(0)
)"'+1.r''+4Y:/1111
:0.
t'(0t:
l),
,:i
til
rvh cre
,ri
I
t'
nt4:l1.i2k 4-lL'
l).r,
-l
[,,.
:/
1._t<
/i l1.j rrJ*I
....
:,i
and0</t<4.
(a) Skctchthegraphof lr(rl.f)bselvethattheareaunderthcgraphisindependentofA.
If il(r) represents a force. this me:rns that the product of the magnitucle o1'the force ancl
thc tinre interval durins rvhich it acts cloes not depend on k.
(b) Write /rttt in tcnls of tlrc unit stcp function ancl thcn solve the given rnitial r,alue
problcnt.
(c)
Plot thc solution lor.
I:
2.
/i
:
l,
n11i1
7,
= *. Dcscribe
horv the solutior-r depends on
A
./ri
i,.{
.;,*a
..
J:il
,.i*
$
,&
:?1,$
.1'',S
tr'il
:tdl
il"{
.silt
rl*l
.ri!
.t:*
348
Chapter 6, The Laplace
Tr
Hon'eveL. if the actual excitation exlends o'er a short, but nonzero. tinrt
then an error will be introducecl by mocleling tire excitation as taking pla;: :.i
taneously, This error rlay be negligible, but in a practical problem it shoi:__ :,
dismissed wrthout consicleration, In problem 16 vou ale iskecl to invesi::,__
issue for a sirnple harrnonic clsciliator.
6'{
PROBLEMS
ln
each of Ploblems 1 through 12:
(a) Fincl the solution of the given initial value problem.
(b) Drau' a graph of the solution.
6e
8L
{.
6,2
6Q
d.2
6a
{L
{L
{.
6a
Ct"c
64,
*
2),' +2),:5(r -:r)l
.1,(0) = 1, ],'(0) = 0
4l: a(r - tr1- 3(t -2r).. l(0) :0. f,(0) :0
3. .t," + 3y' +2y: d(r - 5) * uro(/); )(0) = 0, y,(01 : tlZ
4. )/' - l: -20d(t - 3); .i.(0) = 1, l'(0): ,5. )," * 2.y'-r 3_r,: sint * d(r - 3.zl; l,(0) :0,
l,(01 : g
6, .)," + 4y: 3(t -,1:z); ),(0) : L,/2, f,'(0) = 0
7. y" +.r, : 3(r - 2:z) cos ri
.y(0) : 0, 1'(0) :
6. l" + 11t:1511 - nl4)., 1:(0) :0, 1,,(0) = e
9. )/' I )':./,,,2(r) * 36(l - 3rl2) - Lrz,Q); y(O) :0,
.y'(0) :0
10. 2y" * .v' + ,11, : ri(r - 116) sinr; ),(.0) 0,
=
l,(0) : 0
Il. y" +2)t +2.v:cosr* 3(t -rl2): .v(0)=0, l,(0) =i)
12. .l:(cr -.y : d(r * 1r;
_1,(0) : O. 1"(0) = 0, ),,,(0) = 0. .y,,,(0) : 0
l3 Clonsicier again the st'stem itt Example 1 of this section. in rvhich an oscillation
1.
.y"
2. )/' +
1
., : i-
b1'a unit impulse at r : .5. Suppose that it rs desired to br.ing the sl,stem to resr;:.,
exactly one cycle-tliat is. rvhen thc response llrs1 r.eturns to equilibrium rno\.:.
pt'sitit e tlile ctrun.
(ii) Determine rhe impulse /ra^(r - /0) that shonld be appliecl to the svstem rn ,-:
accornplish this objective. Note that k is tl.re rnagnitude olthe in.rpulse ald rx i. ::.,
of its application.
(b) Solve the resulting initral value problem, and plot its solution to confir:.
behaves in the specilied ntanner.
6a
14. Consider the initial value probleur
\"11t1"+r:d(I-L).
)'(0)
: 0,
y'(0) =
()
where y rs the clantpinq coelficient (or t.e srstunct,).
(a) Let y: ;. Find the solution of the initiar value probrem ancl plot its graph.
(b) Find tl'rc time /t at whicll the solution attains its maximum value. Also :.:
rnaxlmurn value 1,r of the solution.
(c) Let y : j and repeat parts (a) and (b).
(d) Detelnrine horv I, and .y, varv as y decleases. what
ru
e$2 t.S
: 0'l
1r.e
the values of /1 and
Consider the initial value proble nr
\' + )i.\' +_I : /id(/ - l),
t't01
: g,
1"1{t1
: !.
.
_:,
349
,.funatrse Functions
r.vhere
k is the magnitude of an intpulse at I :
resrstance
).
(a) Let
:
value
1'
1. ancl y rs
the damping cocfficient (or
t-.
{. Firrd the value of /r for u,hich the fesponse has a peak value of 2t call this
/c1.
(b) Repeat part (a) for y : t.
(c) Determine horv /<1 varies as 7 decreases. What
62 rc, Consrder
is tl.re value of /ci when ),
:
0?
the initial r'alue problem
+) :.fi(r).
= 0,
tta-1,(t)li2k rvith 0 < k < 1.
)"
_i,(0)
-r,'(0)
=
0,
li(t) : [aa-r(t) (a) Find tlie solution )) : O(.t, k t ol the initral value problem.
(b) Calculatc
p,,. k r flonr tlrc. solution found in palt (a).
wherc
*l-Tr_
(c)
problern
firn frl:3(t
- 4). Find the solulion O0(t) of the given
\\'ith fi(t) replaced hv 6(r - 4).Is it tfue that QoU) :
kllli d(/,k)?
Observe that
(d) Plot 0ft,112), O0,Il4), and dro(t) on the same
rl(t,k) and 0a(t).
Problems 17 through 22 deal rvith the eifect
oscillator'. Suppose that
r,'' + r,
: l'6).
.r
initial value
axes. Describe the relation betrveen
of a sequence ol irnpulses on an undamped
(0)
: 0.
l,'(0) =
0
For each of the lbllon,ing chorccs for f (t):
(a) Trl to predict the nalurc of tl.re solution',vithout solving the problem.
(b) Test 1"our prediction b-v linding the solution and cirarving its graph.
(c) Detelmine
rvhat happens aftel the sequcnce ol irnpulses ends,
2t)
5;L tt. tttt :lArr_
lrrt
L'
d(.-. '.'.
lR ,i,rr:
,,qtQ
)t)
6Q- |e. fQ)
:L6(/-lirl2|
{t,
:
zt. /trt
0, Zl
t5
Eutr
-
(2k
t0
f r-ir'-.5.t-knr
(1,' -,
t(l
CQ
20. f ,r,:
6L
zz lo) : I(-1)i+16(r
l',-lr(
ar1
*
l<tr,2t
-
11kl4)
.1(l
- rttrl
The position of a ccrtaiu lightlv damped oscillator satisfles the initial valuc problettt
-1,,,:0.1-r,,+
v: ft-Dr*,;i, - Lrt, r(0):0.
C)bserve that. except for tl.re danrpirlg
te
,1,'(0):0
nlt, this problem is thc same as Pr oblem
l 8.
(a) Try to predict the nature of the solution rvithout solving the problem
(b) Test your prediction b1' finding the solution and drarving its graph.
(c) Determine what l.rappens
6a- Z+ Ploceecl
as
after the sequence of iupulses
etrcls.
in Problenr 23 for the oscillator satisf-ving
li
j"'*0.1,v'+t':It1r-tl/r-
lr.rl.
-r(0)
:tl.
l'(0) :i)
Observe that. except tor the damping term. this ploblern is tllc same as Probletn 21.
Chapter 6. The Laplace
354
rvhere
ry'(I): I-i{Q(.i)}
ancl
Tt
V/(t):4-L1V1r;}. Obsen'e that r/(r) is the:
the rnrtial value problem
o1,"
a 6l
r
c).'
: Q.
,y(0)
: y9,
y'(0)
:
y'i,,
obtained fron-r Eqs. (20) and (21) by setting g(r) equal to zero. Similarlr.
solution of
st/' + b)i 1cy =
g(t),
: 0.
y(0)
1;'(0)
:
0.
rn rvhich the initial values )6 ancl _vi, are each replacecl by zero.
Once specilic values of a, b. and.' are given. we can frnd rb(.t) : t-L{Ots .'
Table 6.2. 1 . possibl-v in conjunction rvith a translation or a paltierl h'actio n t . : --:
ftr find tttl: t '{q(.r)).it is convenient to write Wis) as
W(s)
:
:
H(s)Gts),
bs * c)-1. The fr.rnction H is known as the transfer
and depends only on the propelties of the systeln under consideration; that :,
determined entirel-v by the coefficients a,b.and c. On the other hand, Gt-r
only on the external excitation g(l) that is applied to the system. By the ccr:.
theorem we can u'rite
rvhere 11(s)
(cs2
+
t-\IH(s'tG(s,]
:
h(t
1,,
-
r)g(r) dr,
rvhere h(.t) : t r{t1(s)}, and g(r) is the given forcing function.
To obtain a better understanding of the signiticance of h(t), we consider t:- =
which G(s) : 1:consequentlv,g(t) : E(t) ancl V(s) : 1/(s). This means tha.
is the solr.rtion of the initial value problem
utt'+by'+c]':r(Il,
.l'(0)
: 0,
),'(0)
:
0,
obtained fron-r Eq. (26) bv replacing g(t) b,v 3(r). Thus /r1r; is the respor:.
svstelr to a unit inpulse applied at /:0, and it rs natural to call lr(t) the
response ol the system, Equation (28) then says that dr(t) is the convolutrr:
inrpulse lespolrstr and the forcing I'urtction.
Referring to Example 2, rve note that in that case, the transfer iur-;,
response is h(t1: (.sin2t)12. Also, the .-terms on the right side of Eq, (19) constitute the function d(t), the solutic,
corresponding homogeneous equation that satisfies the given initial condit: - :
H(s): Il$2 + 4) and the impulse
eh
PROBLEMS
l.
Establish the comnrulative, distributive, ancl associrtn,e propertic-s ol the cc':
integral.
(a)./'iB:g*t
(b) I * (gj +g:) : l *gr - f
(c) l*@*h1:(f *g)'+h
* g,
)'fhis terminologl, alises lrorr tire fact that H(s) is the ratio of the transforms of the oLrtlul on:
of the problem (26).
355
ialr,tolution Integtal
2.Findanexampledifferentfrol-r-rtheoneinthetextshowingthat(fx1)(t)neednotbe
equalto/tt).
-1. Show.
by means of the exampl e -f (t)
:
sin r'
that f + 'f is not necessarilv nonnegatlve'
of the given function'
In each of Problems 4 through T, find the Laplace transform
1, f (t)
=
6. f Q)
:
ln'
t'
- ,;' coslt tlr
{'
-
1,,'
/(r) = Io ,-u-n
5.
'1.
,)r' n,
.f(t')
=/'
sin r dz
- t) cost tlt
sinil
Laplace transform of the given function by
1n each of Problems 8 through 1 1. find the inverse
.rsing the convolution theorem.
S. F-1t) :
O
1
-r.). F(.r)
.1.
11, F(s)
(s+1)2(.s2+4)
(a) If /(1)
s
F, "\
7 r\Jr
I
--,-;---:;
s"{s'+ rJ
= l"' and g(l)
:
1", rvhere n't and
(s+1rts2+4)
G(s)
= -;-:1
s-+r
n are positive integers' show that
I * g- = Y'*"' Jo['
u"'(I
-
u)" dtt'
(b) Llse the convolution theorem to show that
1i
Jo
LL,,{1.
_
tt)" du
=
ft1t tt'.
6 +;T|)..
(c)Exteridthelesultofpart(b)tothecasewheremandnarepositivenumbersbutnot
necessarilv rntegers,
in each of Problems
13 through 20, express the solution
of the given initial vaiue problem in
:erms of a convolution integral'
g(t); y(0) : 0, )'(0) :
0
,J. -\"'+ 21/ +2V: sincv/; y(0) = 0, l'(0) =
i:. ,1r," *4.y'* 17.y = g(l); .v(0) = 0, )'(0) = 0
.6. ,v'' +i'+;y - 1 - uo(I); .v(0) = 1' )'(0) = -1
l,i. y" +,1y' * 4-v : g(1)l )'(0) = 2, "1"(0) = -3
i8, 1," + 3y" + 2.r' : cosa/i -v(0) = i, y'(0) = u
0' y"'(0) = 0
19. y(a, - -'- : g(I); .r-(0) = 0, )"(0) = 0. 1"'(0) =
0, )"(0) :0, )"'(0) = 0
10. ,v(a) + 5.v,'+ 41, : g(r)l y(0) = l, 1,'(0) =
,-r.
-y"
*
a21,
1
=
11. Consider the equatlon
oa') +
ll
Ja
k(t
-
{)d(6) dE = f ft),
inwhlchfandkareknownfuncttons,arrd@istobedetermined.sincetheunknown
called an integra| equation;
given equation is
function @ appears uno., un integral sign. the
known asVolterra integral equations'
eqiations
integral
of
a
class
in particular. it belongs to
and obtain an expression for
-t-ake
the Laplace tr-Jnsform of the girien integral equation
given functions / and k' The
the
of
Lto(/]\in terms of rhe transforms c{l (r)} unic{tr,li
the original integral equatlon
inverse rransform oi C{,lfrl) is the soiution of